Given the function f (x) = (LNX + a) / X (a ∈ R) (1), find the monotone interval of function f (x) (2) when f (x) ≤ 1 Given function f (x) = (LNX + a) / X (a ∈ R) (1) Find the monotone interval of function f (x) (2) when f (x) ≤ 1, find the value range of real number a

Given the function f (x) = (LNX + a) / X (a ∈ R) (1), find the monotone interval of function f (x) (2) when f (x) ≤ 1 Given function f (x) = (LNX + a) / X (a ∈ R) (1) Find the monotone interval of function f (x) (2) when f (x) ≤ 1, find the value range of real number a

1. (0, 1-A power of E) monotonically increasing
[1-A power of E, positive infinity) monotone decreasing
2. LNX + a ≤ x is constant
That is, a ≤ x-lnx is constant
Let g (x) = x-lnx
Derivative of G G '(x) = 1-1 / X
When 00, G (x) is an increasing function
So when x = 1, G (x) has a minimum value of 1
So a ≤ 1
That is, the value range of a is (negative infinity, 1]

The function FX = LNX KX has extremum in 0,1, and the value range of real number k is obtained

f'x=1/x-k
If FX has extremum in 0,1, then f'x = 0 has solution in 0,1, so 1 / x-k = 0 has solution in 0,1
k=1/x
Because 1 / x > 1
So k > 1

ln2-x＞ln(2-x)

According to the meaning of the title, X is less than 2
When x = 0, both sides are equal,
Let C = ln2-x, u = ln (2-x)
From the derivative, we know that C and u are decreasing functions
Let C = 0 get x = LN2, u = 0 get 1, LN2 is less than 1. Draw a graph to know that the interval C where x is less than 0 is above u, so the solution of this inequality is
X is less than 0

Prove the inequality 1 / LN2 + 1 / Ln3 + 1 / ln4 + +1/ln(n+1)
Correction: π / 2 in the question should be n / 2

Consider the function f (x) = 2x / (x + 2) - ln (1 + x)
There are f '(x) = 4 / (x + 2) &# 178; - 1 / (1 + x) = - X & # 178; / ((x + 2) &# 178; (x + 1))
When x > 0, f '(x) < 0, so f (x) decreases strictly monotonically
If f (x) < f (0) = 0, then 2x / (x + 2) < ln (1 + x)
So for any positive integer k, 1 / ln (1 + k) < (K + 2) / (2k) = 1 / 2 + 1 / K
The sum of K from 1 to n is 1 / ln (2) + 1 / ln (3) +... + 1 / ln (n + 1) < n / 2 + 1 + 1 / 2 +... + 1 / n